Answer:
0.2 m/s^2
Explanation:
initial speed 14m/s
final speed 20m/s
acceleration:
(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2
Answer:
8.85m/s
Explanation:
The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.
When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.
We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s
The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
- mass of the toy, m = 0.1 kg
- the maximum height reached by the, h = 0.45 m
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E
Substitute the given values and solve the speed;
Thus, the speed of the toy when it hits the ground is 2.97 m/s.
Learn more here: brainly.com/question/7562874
I can answer it in the comments, but can i ask what i’m supposed to do exactly? i remember learning about this stuff but i need the instructions.
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
